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3e-spring2011-chapter_2_tf

# 3e-spring2011-chapter_2_tf - Chapter 2 Ch 2.TF 7 T by...

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Unformatted text preview: Chapter 2 Ch 2.TF. 7 T, by Theorem 2.2.4. Ch 2.TF. 8 T, by Theorem 2.4.6. Ch 2.TF. 9 T; The matrix is bracketleftbigg 1 − 1 − 1 1 bracketrightbigg . Ch 2.TF. 10 F; The columns of a rotation matrix are unit vectors; see Theorem 2.2.3. Ch 2.TF. 11 F; Note that det( A ) = ( k − 2) 2 + 9 is always positive, so that A is invertible for all values of k . Ch 2.TF. 12 T; Note that the columns are unit vectors, since ( − . 6) 2 + ( ± . 8) 2 = 1. The matrix has the form presented in Theorem 2.2.3. Ch 2.TF. 13 F; Consider A = I 2 (or any other invertible 2 × 2 matrix). Ch 2.TF. 14 T; Note that A = bracketleftbigg 1 2 3 4 bracketrightbigg − 1 bracketleftbigg 1 1 1 1 bracketrightbiggbracketleftbigg 5 6 7 8 bracketrightbigg − 1 is the unique solution. Ch 2.TF. 15 F, by Theorem 2.4.9. Note that the determinant is 0. Ch 2.TF. 16 T, by Theorem 2.4.3. Ch 2.TF. 17 T; The shear matrix A = bracketleftbigg 1 1 2 1 bracketrightbigg works. Ch 2.TF. 18 T; Simplify to see that T bracketleftbigg x y bracketrightbigg = bracketleftbigg 4 y − 12 x bracketrightbigg = bracketleftbigg 4 − 12 bracketrightbiggbracketleftbigg x y bracketrightbigg . Ch 2.TF. 19 T; The equation det( A ) = k 2 − 6 k + 10 = 0 has no real solution. Ch 2.TF. 20 T; The matrix fails to be invertible for k = 5 and k = − 1, since the determinant det A = k 2 − 4 k − 5 = ( k − 5)( k + 1) is 0 for these values of k . Ch 2.TF. 21 T; The product is det( A ) I 2 . Ch 2.TF. 22 T; Writing an upper triangular matrix A = bracketleftbigg a b c bracketrightbigg and solving the equation A 2 = bracketleftbigg bracketrightbigg we find that A = bracketleftbigg b bracketrightbigg , where b is any nonzero constant....
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3e-spring2011-chapter_2_tf - Chapter 2 Ch 2.TF 7 T by...

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